\newproblem{lay:1_7_39}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 1.7.39}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Suppose $A$ is a $m\times n$ matrix with the property that for all $\mathbf{b}$ in $\mathbb{R}^m$, the equation
	$A\mathbf{x}=\mathbf{b}$ has at most one solution. Use the definition of linear independence to explain why 
	the columns of $A$ must be linearly independent.
}{
  % Solution
  If $A\mathbf{x}=\mathbf{b}$ has at most one solution for all $\mathbf{b}$, then in particular for $\mathbf{b}=\mathbf{0}$, the
	equation $A\mathbf{x}=\mathbf{0}$ has at most one solution. But we already know that $\mathbf{x}=\mathbf{0}$ is a solution (the trivial solution).
	So it must be its only solution. Let us refer to the columns of $A$ as $\mathbf{a}_i$ ($i=1,2,...,n$). The equation $A\mathbf{x}=\mathbf{0}$ can be
	rewritten as
	\begin{center}
		$x_1\mathbf{a}_1+x_2\mathbf{a}_2+...+x_n\mathbf{a}_n=\mathbf{0}$
	\end{center}
	Because the trivial solution is its only solution, then the set of column vectors of $A$ is linearly independent.
	
}
\useproblem{lay:1_7_39}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
